The frequency of revolution of an electron revolving in a circular path in a magnetic field of \( 7.8 \times 10^{-4}\,\mathrm{T} \) will be (Take \( e = -1.6 \times 10^{-19}\,\mathrm{C};\ m_e = 9.1 \times 10^{-31}\,\mathrm{kg};\ \pi = 22/7 \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 21.8 MHz
In a uniform magnetic field the magnetic force \( |q|vB \) provides the centripetal force for the circular motion.
\( \displaystyle \frac{m v^{2}}{r} = |q|\,v B \;\Rightarrow\; r = \frac{m v}{|q| B} \)
\( \displaystyle f = \frac{v}{2\pi r} = \frac{|q|\,B}{2\pi m} \)
Only the magnitude of the charge enters; the sign decides the sense of revolution.
\( \displaystyle f = \frac{1.6\times10^{-19} \times 7.8\times10^{-4}}{2 \times \frac{22}{7} \times 9.1\times10^{-31}} = \frac{1.248\times10^{-22}}{5.72\times10^{-30}} \)
\( \displaystyle f = 2.18\times10^{7}\,\mathrm{Hz} = 21.8\,\mathrm{MHz} \)